Problem 47 - Algebra A Combined Function

Question

It's the end of the budgeting period for Dennis Fernandes, and he has \(\$ 500\) left in his budget for car rental expenses. He plans to spend this budget on a sales trip throughout southern Texas. He will rent a car that costs \(\$ 30\) per day and \(\$ 0.15\) per mile and he can spend no more than \(\$ 500\). a. Write an inequality describing this situation. Let \(x=\) number of days and let \(y=\) number of miles. b. Graph this inequality below. C. Why is the grid showing quadrant I only?

Step-by-Step Solution

Answer

a. Inequality: \(30x + 0.15y \leq 500\). c. Only quadrant I is shown because x and y cannot be negative.

Step 1: Understanding the problem

Dennis has $500 available for car rental expenses, with a daily cost of $30 and a per mile cost of $0.15. We want to find how he can spend this budget without exceeding $500.

Step 2: Formulate the inequality

Let \( x \) represent the number of days Dennis rents the car and \( y \) represent the number of miles he drives. The total cost can be represented as \(30x + 0.15y\). Since Dennis cannot spend more than $500, the inequality is \(30x + 0.15y \leq 500\).

Step 3: Graph the inequality

To graph the inequality \(30x + 0.15y \leq 500\) in the coordinate plane, rearrange it to find \( y \): \[0.15y \leq 500 - 30x\] Divide by 0.15: \[ y \leq \frac{500 - 30x}{0.15}\] This equation can be graphed starting from y-intercept (when x=0). The inequality will show the area below and including the line, restricting to quadrant I.

Step 4: Why Grid Shows Quadrant I Only

In this context, both \( x \) and \( y \) must be non-negative because they represent the number of days and miles, respectively, both of which cannot be negative. This confines the realistic solutions to quadrant I, where both x and y are positive.

Key Concepts

Budgeting in Mathematics

Budgeting is all about planning how to spend a given amount of money without exceeding it. In mathematics, this often involves setting up equations or inequalities to represent financial scenarios.
A budget, like the one Dennis is managing, requires careful consideration of different expenses.

• Firstly, identify fixed costs, such as the daily rental rate of the car at $30 per day.
• Then, include variable costs, which in this case, are based on usage like mileage, costing $0.15 per mile.
• Next, sum these costs to form a total expense that should not exceed your budget limit, here represented as an inequality.

The inequality helps us understand the limits within which Dennis can operate, ensuring he does not exceed his budget, thereby allowing successful financial planning. This type of budgeting is essential in numerous real-life situations, like planning a family vacation, business trips, or even monthly expenses.

Graphing Inequalities

Graphing inequalities is a powerful way to visually represent constraints in mathematical problems. In Dennis's scenario, the inequality is vital in determining his constraints for rental expenses.
To graph

• Start by arranging the inequality into the form that easily suggests how to sketch the line, here done as: \[ y \leq \frac{500 - 30x}{0.15} \]
• The line itself symbolizes the boundary of the budget's constraint, while the shaded region below represents all possible combinations of days and miles Dennis can afford.
• To draw the graph, identify the y-intercept (where x=0), and use this to plot the initial point. Then, find the slope to extend the line throughout the coordinate plane emphasizing quadrant I.

The final graph informs Dennis of all possible days and mileage combinations that will keep him within the $500 budget. It's not just a graph but a tool to make informed decisions.

Coordinate Plane

The coordinate plane is critical for visually representing relationships between variables in mathematics. In problems like Dennis's, it inspects the area of possible solutions, constrained by inequalities.
The usage of only Quadrant I, where both x (days) and y (miles) must remain non-negative, is straightforward:

• Quadrant I is essential when variables cannot be negative, which is the case in many real-world scenarios involving time and distance.
• Here, every point in Quadrant I signifies a feasible solution, underlining both practical and mathematical correctness.
• It prevents impractical results, like negative days or miles, ensuring only logical outcomes are considered.

Finally, this ensures that someone's plan, like Dennis's, follows practical constraints, crafting a graphical representation that aligns with real world possibilities.